Chapter 4. square sum graphs. 4.1 Introduction

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1 Chapter 4 square sum graphs In this Chapter we introduce a new type of labeling of graphs which is closely related to the Diophantine Equation x 2 + y 2 = n and report results of our preliminary investigations on this new concept. Let G be a (p, q)-graph. G is said to be a square sum graph if there exist a bijection f : V (G) {0, 1,..., p 1} such that the induced function f : E(G) N given by f (uv) = [f(u)] 2 + [f(v)] 2 for every uv E(G) are all distinct. The square sum labeling f is called a prime square sum labeling if f (uv) is 1 or a prime number uv E(G). We prove that for a prime square sum graph G, f(u) and f(v) are relatively prime e = uv E(G) with f (e) 1. Moreover we prove that the complete graph K n for n 5, the cycles C n for n 3, trees, cycle cactus, ladders and the complete lattice grids are square sum. We also proved that every graph can be embedded into a square sum graph. Also we characterize the class of prime square sum graphs. 4.1 Introduction Abundant literature exists as of today concerning the structure of graphs admitting a variety of functions assigning real numbers to their elements so that certain given conditions are satisfied. Here we are interested the study of vertex functions f : V (G) A, A N for 78

2 Chapter 4 79 which the induced edge function f : E(G) N is defined as f (uv) = [f(u)] 2 + [f(v)] 2, e E(G) are all distinct. As we know that the notion of prime labeling originated with Entrnger and was introduced in a paper by Tout, Dabboucy and Howalla [Gal05]. A graph with vertex set V is said to have a prime labeling if its vertices are labeled with distinct integers 1, 2,..., V such that for each edge xy the labels assigned to x and y are relatively prime [Gal05]. This concept leads to define a prime square sum graph which is a square sum graph with edge values are only prime numbers together with 1. Our motivation is to study how the number theoretic problems are related to the structure of graphs. Definition Let G be a (p, q)-graph. G is said to be a square sum graph if there exist a bijection f : V (G) {0, 1,..., p 1} such that the induced function f : E(G) N given by f (uv) = [f(u)] 2 + [f(v)] 2 for every uv E(G) are all distinct. The square sum labeling f is called a prime square sum labeling if f (uv) is 1 or a prime number uv E(G). Figures 4.1 and 4.2 give two graphs which are square sum and two graphs which are prime square sum respectively. Figure 4.1:

3 Chapter 4 80 Figure 4.2: The following theorem is a consequence of the definition of a square sum graph. Theorem For any (p, q)-graph G = (V, E) and for any square sum labeling f : V (G) {0, 1, 2,..., p 1}, we have f (e) = [f(u)] 2 d(u) e E(G) u V (G). Proof. Let G be a square sum graph. Then f (e i ) = [f(u i )] 2 +[f(v i ) 2 ] for all e i = (u i v i ) E(G) so that f (e) = [f(u i )] 2 + [f(v i ) 2 ]. In the sum f (e) each f(u i ) occurs d(u i ) times. Hence f (e) = [f(u)] 2 d(u) e E(G) u V (G). Corollary If G = (V, E) be a (p, q)-graph which is r-regular and square sum, then 6 f (e) = rp(p 1)(2p 1).

4 Chapter 4 81 Proof. By the theorem 4.1.2, f (e) = [f(u)] 2 d(u) e E(G) u V (G). Since G is r-regular, d(u i ) = r for all u i V (G). Hence f (e) = r [f(u)] 2 = rp(p 1)(2p 1) 6. Theorem Let G = (V, E) be a (p, q)-graph which is square sum graph with the square sum labeling f. Then f (e) can be represented as the sum of two squares. Proof. Let G be a graph having two edges say e 1 = u 1 v 1 and e 2 = u 2 v 2. Then f (e 1 ) = [f(u 1 )] 2 +[f(v 1 )] 2 and f (e 2 ) = [f(u 2 )] 2 +[f(v 2 )] 2. Let f(u 1 ) = a, f(v 1 ) = b, f(u 2 ) = c and f(v 2 ) = d so that f (e 1 ).f (e 2 ) = (a 2 +b 2 ).(c 2 +d 2 ) = (ac+bd) 2 +(ad bc) 2. By applying the principle of induction on the number of edges we get the result. As we know that certain numbers like 3, 6, 7 etc.cannot be written as sum of two squares. Hence there are forbidden numbers which cannot appear as an edge labeling. Hence the interesting question is, what are those numbers which are not forbidden as edge labeling. We have the following theorem. Theorem Let G be a square sum graph which admits a square sum labeling f. Then f (e) not congruent to 3(mod 4), e E(G).

5 Chapter 4 82 Proof. If possible, let f (e) 3(mod 4) for some e E(G) Let f (e) = a 2 + b 2 where e = (uv), f(u) = a and f(v) = b. Then a 2 + b 2 3(mod 4). This is not possible. Hence f (e) not congruent to 3(mod 4), e E(G). Theorem Let G be a square sum graph which admits a square sum labeling f. If f (e) 1(mod2), e E(G), then G is bipartite. Proof. Let us suppose that f (e) 1(mod 2), e E(G) Let f (e) = a 2 + b 2 where e = uv E(G), f(u) = a, f(v) = b. Then a 2 + b 2 1(mod 2). Since a 2 + b 2 is an odd number, we have a and b are of opposite parity. Let X = {u : f(u) is even } and Y = {v : f(v) is odd }. The sets X and Y form a disjoint partition of V (G) and any edge with f (e) 1(mod 2) is of one end in X and the other end in Y. proposition The following two cases are not possible for a square sum graph. (i)f (e) 2(mod 4), e E(G) (ii)f (e) 0(mod 4), e E(G). Proof. (i) Suppose for all edge e = uv E(G). f (e) 2(mod 4). But f (e) = [f(u)] 2 + [f(v)] 2 2(mod 4). Let f(u) = 0. Then [f(v)] 2 2(mod 4), which is not possible since the square of an integer is either congruent to 0(mod 4) or congruent to 1(mod 4). Hence 0 does not belong to f(v (G)). (ii) Suppose for all edge e = uv E(G) we have f (e) 0(mod 4).

6 Chapter 4 83 Hence f (e) = [f(u)] 2 + [f(v)] 2 0(mod 4). Let f(u) = 1. Then [f(v)] 2 1( mod 4), which is not possible since the square of an integer is either 0(mod 4) or 1(mod 4). Hence 1 does not belong to f(v (G)). proposition If f (e) 1(mod 4), e E(G) G is bipartite. Proof. f (e) 1(mod 4), e E(G) f (e) 1(mod 2), e E(G). Then apply Theorem. 4.2 Some classes of square sum graphs Theorem The graph G = K 2 + mk 1 is a square sum graph. Proof. Let V (G) = {u 1, u 2,...u m+2 } where V (K 2 ) = {u 1, u 2 }. Define f : V (G) {0, 1,..., m + 1} by f(u i ) = i 1, 1 i m + 2. Clearly, the induced function f is injective, for if f (u 1 u i ) = f (u 2 u j ), then we get [f(u 1 )] 2 + [f(u i )] 2 = [f(u 2 )] 2 + [f(u j )] 2. By taking f(u i ) = x and f(u j ) = y, we get an equation of the form x 2 y 2 = 1, which implies either one of x or y is zero. That is either f(u i ) = 0 or f(u j ) = 0, which is not possible since f is a bijection. Hence G is a square sum graph.

7 Chapter 4 84 The following figure 4.3 illustrates the theorem for m = 4. Figure 4.3: Theorem and only if n 5. The complete graph K n is a square sum graph if Proof. The square sum labeling of the complete graph K n for n 5 is given in figures?? and??. Now, let K n, n 6 and f : V (G) {0, 1, 2,..., n 1} be a vertex function which induces a function f given by f (xy) = [f(x)] 2 + [f(y)] 2. Assume n 6 and since the graph is a complete graph we get two edges e 1 and e 2 such that f (e 1 ) = = 25 and f (e 2 ) = = 25. Hence, we conclude that when n 6, f is not injective Hence K n, n 6 is not a square sum graph. Theorem Cycles are square sum graphs. Proof. Let C p be a cycle of length p and let C p = (u 1 u 2...u p u 1 ). Case(i): p is odd. Define f : V (C p ) {0, 1,..., p 1} as f(u i ) = i 1, 1 i p. Here f is an increasing function on V (C p ), so f is also an increasing function on E(C p ) {u p u 1 }. Hence f (e i ) f (e j ), i j, for every e i, e j E(C p ) {u p u 1 }. When p is odd, f(u p ) is even. Therefore f (u p u 1 )

8 Chapter 4 85 is even and hence f (u p u 1 ) f (e j ) for every e j E(C p ) {u p u 1 }; hence f is injective and f is a square sum labelling on C p. Case(ii): when p is even and p 6. We know that the equation a 2 + (a + 1) 2 = c 2 has only one non-zero positive consecutive integer solution which are given by a = 3, a+1 = 4 and c = 5. Hence for C p, p 6 we can have the same labeling as in Case (i). For C p, p = 6, consider the following particular labeling given by, f : V (C p ) {0, 1, 2, 3, 4, 5} as f(u i ) = i 1, 1 i 4, f(u 5 ) = 5 and f(u 6 ) = 4. Then f is injective and f is a square sum labeling of C 6. The square sum labeling of cycles C 5 and C 6 are given in the figure 4.4. Figure 4.4: Theorem Trees are square sum graphs. Proof. Let T be a tree and let v 1 V (T ) be the vertex with maximum degree. Start from the vertex v 1, and applying BFS algorithm, label the vertices of T with 0, 1, 2,..., p 1 in the order in which they are visited. Since f is increasing on the vertex set of T and

9 Chapter 4 86 f (u i u j ) = [f(u i )] 2 + [f(u j )] 2, we have f is also an increasing function on the edge set of T. Hence f is injective and f is a square sum labeling on T. Remark Spanning subgraph H of a square sum graph G is square sum since V (G) = V (H) and f(v (G)) = f(v (H)) = {0, 1,..., p 1}. But all subgraphs of a square sum graph need not be square sum as the following figure 4.5 illustrates. Figure 4.5: Definition A cycle-cactus is a graph which consisting of n copies of C k, k 3 concatenated at exactly one vertex is denoted as C (n) k. Theorem The cycle-cactus C (n) k is a square sum graph. Proof. For a given integer k 3, let C (n) k be the cycle-cactus consisting of n copies of the cycle C k, denoted G 1, G 2,..., G n, all concatenated at exactly one vertex, say z. Let the vertices of G i, 1 i n be labelled z, u i1, u i2,..., u i(k 1), 1 i n. Define f : V (C (n) k )

10 Chapter 4 87 {0, 1,..., p 1}. First of all, let f(z) = n(k 1). Case (i) k is odd. Label the two antipodal vertices of the cycle G 1 by 1 and 2 respectively, of the cycle G 2 by 3 and 4 respectively,... and label the two antipodal vertices of the cycle G n by 2n 1 and 2n respectively. Next, assign the number 2n + 1 to the unlabeled vertex adjacent to the vertex labeled 1 in G 1, 2n + 2 to the unlabeled vertex adjacent to the vertex labeled 2 in G 1, 2n + 3 to the unlabelled vertex adjacent to the vertex labeled 3 in G 2, 2n + 4 to the unlabeled vertex adjacent to the vertex labeled 4 in G 2,..., 4n 1 to the unlabeled vertex adjacent to the vertex labeled 2n 1 in G n, and 4n to the unlabeled vertex adjacent to the vertex labeled 2n in G n. Next, assign the number 4n+1 to the unlabeled vertex adjacent to the vertex labeled 2n + 1 in G 1, 4n + 2 to the unlabeled vertex adjacent to the vertex labeled 2n+2 in G 1, 4n+3 to the unlabeled vertex adjacent to the vertex labeled 2n+3 in G 2, 4n+4 to the unlabeled vertex adjacent to the vertex labeled 2n + 4 in G 2,..., 6n 1 to the unlabeled vertex adjacent to the vertex labeled 4n 1 in G n, and 6n to the unlabeled vertex adjacent to the vertex labeled 4n in G n. Continuing in this manner, we end up assigning the number (k 3)n+1 to the unlabeled vertex adjacent to the vertex labeled (k 5)n + 1 in G 1, (k 3) + 2 to the unlabeled vertex adjacent to the vertex labeled (k 5)n + 2 in G 1, (k 3)n+3 to the unlabeled vertex adjacent to the vertex labeled (k 5)n+3 in G 2, (k 3)n+4 to the unlabeled vertex adjacent to the vertex labeled (k 5)n + 4 in G 2,..., (k 1)n 1 to the unlabeled

11 Chapter 4 88 vertex adjacent to the vertex labeled (k 3)n 1 in G n, and (k 1)n to the last unlabeled vertex adjacent to the vertex labeled (k 3)n in G n. Case (ii) k is even. In this case, label the unique antipodal vertex of the cycle G i by i, 1 i n. Next, assign the number n + 1 and n + 2 to the unlabeled vertices adjacent to the vertex labeled 1 in G 1, n + 3 and n + 4 to the unlabeled vertices adjacent to the vertex labeled 2 in G 2, n + 5 and n + 6 to the unlabeled vertices adjacent to the vertex labeled 2 in G 2, n + 5 and n + 6 to the unlabeled vertices adjacent to the vertex labeled 3 in G 3, n+7 and n+8 to the unlabeled vertices adjacent to the vertex labeled 4 in G 4,..., 3n 1 and 3n to the unlabeled vertices adjacent to the vertex labeled n in G n. Next, assign the number 3n + 1 to the unlabeled vertex adjacent to the vertex labelled n+1 in G 1, 3n+2 to the unlabeled vertex adjacent to the vertex labeled n + 2 in G 1, 3n + 3 to the unlabeled vertex adjacent to the vertex labeled n+3 in G 2, 3n+4 to the unlabeled vertex adjacent to the vertex labeled n + 4 in G 2,..., 4n 1 to the vertex adjacent to the unlabeled vertex labeled 3n 1 in G n, and 4n to the unlabeled vertex adjacent to the vertex labeled 3n in G n. Next, assign the number 4n+1 to the unlabeled vertex adjacent to the vertex labeled 3n+1 in G 1, 4n + 2 to the unlabeled vertex adjacent to the vertex labelled 3n + 2 in G 1, 4n + 3 to the unlabeled vertex adjacent to the vertex labeled 3n + 3 in G 2, 4n + 4 to the unlabeled vertex adjacent to the vertex labeled 3n + 4 in G 2,..., 6n 1 to the vertex adjacent to the unlabeled vertex labeled 4n 1 in G n, and 6n to the unlabeled vertex

12 Chapter 4 89 adjacent to the vertex labeled 4n in G n. Continuing in this manner, we end up assigning the number (k 3)n + 1 to the unlabeled vertex adjacent to the vertex labeled (k 5)n+1 in G 1, (k 3)+2 to the unlabeled vertex adjacent to the vertex labeled 2n+2 in G 1, (k 3)n+3 to the unlabeled vertex adjacent to the vertex labeled (k 5)n + 3 in G 2, (k 3)n + 4 to the unlabeled vertex adjacent to the vertex labeled (k 5)n + 4 in G 2,..., (k 1)n 1 to the unlabeled vertex adjacent to the vertex labeled (k 3)n 1 in G n, and (k 1)n to the last unlabeled vertex adjacent to the vertex labeled (k 3)n in G n. In the above two cases we have f(u 11 ) < f(u 21 ) < f(u 31 ) <... < f(u n1 ). f(u 12 ) < f(u 22 ) < f(u 32 ) <... < f(u n2 ). f(u 13 ) < f(u 23 ) < f(u 33 ) <... < f(u n3 ) f(u 1(k 1) ) < f(u 2(k 1) ) < f(u 3(k 1 ) <... < f(u n(k 1) ) and f(z) = n(k 1). Hence the induced function on the set of edges of C (n) k is injective so that the assignment f defined above in each case is a square sum labeling of C (n) k. Corollary The friendship graph C (n) 3 is a square sum graph. Proof. Let k = 3 in the above theorem.

13 Chapter 4 90 Corollary For the star K 1,p 1, if we label the central vertex with any integer between 0 and p 1 and the remaining values are assigned to the vertices of unit degree in a one to one manner, we get a square sum labeling. Theorem Ladder L n is a square sum graph. Proof. Let V (L n ) = {a 1, a 2,..., a n, b 1, b 2,..., b n } and E(L n ) = {a i b i : 1 i n} {a i a i+1 : 1 i n 1} {b i b i+1 : 1 i n 1}. Define f : V (L n ) {0, 1..., p 1} as follows: f(a i ) = 2i 2, 1 i n and f(b i ) = 2i 1, 1 i n so that f (a i b i ) = (2i 2) 2 + (2i 1) 2, 1 i n. Next, f (a i a i+1 ) = (2i 2) 2 + (2i) 2, 1 i n 1, f (b i b i+1 ) = (2i 1) 2 + (2i + 1) 2, 1 i n 1; since f (a i b i ) is an odd number for 1 i n and f (a i a i+1 ).Also f (b j b j+1 ), 1 i n 1, 1 j n 1 are both even numbers, we have f (a i b i ) f (a i a i+1 ) and f (a i b i ) f (b i b i+1 ). Moreover, f is an increasing function on V (G) and so f (a i b i ) f (a j b j ), i j and f (a i a i+1 ) f (a j a j+1 ), i j, 1 i n 1, 1 j n 1. Also, f (b i b i+1 ) f (b j b j+1 ) and f (a i b i ) f (b j b j+1 ), 1 i n 1, 1 j n 1, since these numbers form an increasing sequence of even numbers. Hencef is injective and f is a square sum labeling.

14 Chapter 4 91 The following figure 4.6 gives a square sum labeling of L n for n = 5. Figure 4.6: Theorem square sum graphs. The complete lattice grids L mn = P m P n are Proof. Let G be the grid with V (G) = {u 11, u 12,..., u 1n, u 21, u 22,..., u 21,..., u m1, u m2,..., u mn }. We arrange the set of vertices of G into different levels as follows: u 11 u 21, u 12 u 31, u 22, u 13,..., u mn. That is V (G) = {u ij : i + j = 2, 3,..., m + n, i m, j n}. Define f : V (G) {0, 1,..., m.n 1}, by f(u 11 ) = 0, f(u 21 ) = 1, f(u 12 ) = 2, f(u 31 ) = 3, f(u 22 ) = 4, f(u 13 ) = 5.

15 Chapter 4 92 After labelling the i th level vertices we will label the (i + 1) th level of vertices and so on. Then the induced edge labeling given by f (e) = [f(u)] 2 + [f(u)] 2, e E(G), is injective, for if let e 1 = u 1 v 1 and e 2 = u 2 v 2 be any two edges of G. (i)u 1 = u 2 f(v 1 ) < f(v 2 ) or f(v 1 ) > f(v 2 ), then f (e 1 ) f (e 2 ). (ii) If u 1 u 2, then f(u 1 ) < f(u 2 ) f(v 1 ) < f(v 2 ). Hence f (e 1 ) f (e 2 ) so that f is a square sum labeling of G. Hence G is a square sum graph. The following figure 4.7 illustrates the theorem for m = 4, n = 5. Figure 4.7: Next we are looking for bipartite graph which are square sum. The following theorem gives the idea that there exists plenty of graphs which are not square sum. Theorem if m 4, n. The complete bipartite graphs K m,n is square sum

16 Chapter 4 93 Proof. Let V (K m,n ) = X Y X Y = φ. Stars K 1,n, n are square sum by theorem [3.9]. The different sets of partitions of the vertex sets of K m,n, 2 m 5 are given below: Case (i) m = 2. X = {x 1, x 2 } and Y = {y 1, y 2,..., y n }, define f(x 1 ) = 0, f(x 2 ) = 1 and f(y i ) = i + 1, 1 i n. Case (ii) m = 3. X = {x 1, x 2, x 3 } and Y = {y 1, y 2,..., y n }, define f(x 1 ) = 0, f(x 2 ) = 1, f(x 3 ) = 2 and f(y i ) = i + 2, 1 i n. Case (iii) m = 4. X = {x 1, x 2, x 3, x 4 } and Y = {y 1, y 2,..., y n }, define f(x 1 ) = 0, f(x 2 ) = 1, f(x 3 ) = 2f(x 4 ) = 3 and f(y i ) = i + 3, 1 i n. consider the following sets of equations. x 2 y 2 = 1, x 2 y 2 = 4, x 2 y 2 = 9, x 2 y 2 = 16, x 2 y 2 = 4, x 2 y 2 = 8, x 2 y 2 = 5, x 2 y 2 = 12, x 2 y 2 = 7. From the solutions of these nine equations we conclude that f is injective for K m,n, m 4, n. Hence K m,n is square sum if m 4, n. Problem Characterize the the class of bipartite square sum graphs. We have already give some graphs which are not square sum. Therefore our next attempt is to embed such graphs into square sum graphs.

17 Chapter 4 94 When we succeed in this effort then try to make embeddings into an eulerian, bipartite, hamiltonian embeddings. 4.3 Embeddings of square sum graphs Theorem Every (p, q)-graph G can be embedded in a connected square sum graph H with 5 p 1 + t + q p edges and 5 p vertices where t is the number of connected components of G. Proof. Let G be a graph with vertex set V (G) and let V (G) = {u 1, u 2,..., u p }. We will embed the graph G in a graph H with V (H) = 5 p and E(H) = 5 p 1 + t + q p where t is the number of connected components of G. Consider the set of p integers 0, 5, 5 2,..., 5 p 1. Label the vertices of G with the above p numbers by f(u 1 ) = 0, f(u i ) = 5 i 1, 2 i p. Introduce isolated vertices v 1, v 2,..., v n where n = 5 p p. Labeled these n isolated vertices by 1, 2, 3, 4, 6, 7,..., 24, 26, 27,..., 124, 126,..., 5 p 1 1. (Note that here we are omitting the numbers 5, 5 2,..., 5 p 1 ). If G is connected join u 1 to all v k, 1 k n. If G is disconnected and having t components, say C 1, C 2,..., C t, join u 1 to exactly one of the vertex of C i, say u i1, 2 i t, where the vertex u 1 belongs to the component C 1. Then for the resulting graph H, f is a bijection from V (H) {0, 1,..., 5 p 1 }. The induced edge labeling f is injective, for if assume there exists edges e 1 and e 2 such that f (e 1 ) = f (e 2 ).

18 Chapter 4 95 Case (i). Let e 1 = u k u l and e 2 = u i u j where 1 i, j, k, l p.if f (e 1 ) = f (e 2 ) which implies (5 i ) 2 + (5 j ) 2 = (5 k ) 2 + (5 l ) 2. (i) Let no end vertex of e 1 and e 2 be labeled as zero and since i, j, k, l 0, divide throughout by (5 i ) 2 where i be the least among them. Hence j i = 5 k i + 5 l i which is a contradiction. (ii) Let one of the end vertex of e i or e j be labeled as zero. Then, f (e i ) = f (e j ) (5 i ) 2 = (5 k ) 2 + (5 l ) 2, again contradiction. (iii) Let one end vertex of e i and e j be labeled as zero. Then f (e 1 ) = f (e 2 ) (5 i ) 2 = (5 k ) 2, which implies i = j, which is not possible. Case (ii). Let e 1 = u 1 v i and e 2 = u 1 v j where 1 i n and 1 j n. Since f(u 1 ) = 0 and if we assume f(v i ) = x i and f (v j ) = x j, then f (e 1 ) = f (e 2 ) f (u 1 v i ) = f (u 1 v j ), which implies that x 2 i = x 2 j, not possible, hence f (u 1 v i ) f (u 1 v j ), i j. Case (iii). Let e 1 = u 1 v i and e 2 = u 1 u k where 1 i n and 1 k n. Then f (e 1 ) = f (e 2 ) x 2 i = (5 k ) 2. Hence we get x i = 5 k, not possible since f is a bijection. Case (iv). Let e 1 = u k u l and e 2 = u 1 v i, then f (e 1 ) = f (e 2 ) (5 k ) 2 + (5 l ) 2 = x 2 i, since the left side is even,x i must be even. Hence x 2 i 0(mod4) but (5 k ) 2 + (5 l ) 2 2(mod4), which is a contradiction. From the above four cases we conclude that the induced function f is a square sum labeling of H and hence the graph H is a square sum graph which contains G as a subgraph.

19 Chapter 4 96 When the given graph is connected the following figure 4.8 represents the embedding of the graph into a connected square sum graph. Figure 4.8: When the given graph is disconnected the following figure 4.9 represents the embedding of the graph into a connected square sum graph. Figure 4.9: Corollary If G is planar, so the square sum graph H.

20 Chapter 4 97 Corollary Every eulerian graph can be embedded into an eulerian square sum graph. Proof. Let G be the given eulerian graph. By the theorem 4.3.1, we can embed G into a square sum graph. In that embedding if we construct distinct cycles joining the vertices with labeling given by Similarly cycles with vertex labeling given by Here each cycle is centered at the vertex with labeling 5. Then the embedding becomes an eulerian embedding. Corollary Every bipartite graph can be embedded into a bipartite square sum graph. Proof. Since each cycle in the corollary centered at the vertex with labeling 5 is of even length we have if the given graph is bipartite then the embedding also bipartite. Corollary In a similar way one can show that very Hamiltonian graph can be embedded into a Hamiltonian square sum graph. 4.4 Prime square sum graphs As we know that prime numbers play an important role in number theory, We investigate the structure of square sum graphs with edge labeling prime numbers. Theorem Let G be a prime square sum graph. Then f (e) 1(mod4), e E(G).

21 Chapter 4 98 Proof. If f (e) = 1, then obviously f (e) 1(mod4). Otherwise assume f (e) 1. Since f (e) is a prime number either f (e) 1(mod4) or f (e) 3(mod4). But by theorem 4.1.5, f (e) cannot be congruent to 3(mod4), hence f (e) 1(mod4). Observation graph. All prime square sum graphs are bipartite The following corollary is immediate. Corollary The number of vertices p of a prime square sum graph is even if and only if X = Y and if the number of vertices p is odd if and only if X Y = 1. Theorem If G is a prime square sum graph, then it will have at least one pendant vertex. Proof. Let G be a prime square sum graph and let the vertex u be such that f(u) = 0. Then the vertex u can be adjacent only with the vertex whose labeling is 1. Hence the degree of the vertex u with f(u) = 0 is always 1. Remark The condition given in Theorem is not sufficient as the following figure 4.10 illustrates.

22 Chapter 4 99 Figure 4.10: Theorem ), e E(G). For any prime square sum graph f (e) 1, 5( mod Proof. Let e = uv E(G). Since G is prime square sum, either f(u) is even and f(v) is odd or vice versa. But the square of an odd integer is always congruent to 1( mod 8) and square of an even integer is congruent to 0, 4(mod 8). Hence [f(u)] 2 + [f(v)] 2 1, 5(mod 8).. Theorem For any prime square sum graph G, f(u) and f(v) are relatively prime for all e E(G) except the edge e = uv with f (e) = 1 Proof. Let G be a prime square sum graph and e = uv E(G) such that f (e) 1.Hence f(u) 0 and f(v) 0. Let us assume f(u) and f(u) are not relatively prime and let d be the greatest common divisor of f(u) and f(v). We have f (e) = f (uv) = [f(u)] 2 + [f(v)] 2 and since d divides both f(u) and f(v), it must divide f (e) also; which is a contradiction to the fact that f (e) is a prime. Hence f(u) and f(v) are relatively prime for all e E(G) except the edge e = uv

23 Chapter with f (e) = 1. Now we give an interesting result. We have already mentioned the definition of a prime graph. A graph G = (V, G) is said to be a prime graph if its vertices are labelled with distinct integers 1, 2,..., V such that for each edge xy, the labels assigned to x and y are relatively prime [?]. The following theorem is an immediate result obtained from theorem Theorem All prime square sum graphs are prime graphs. Proof. Let G be a prime square sum graph. Then there exists a function f : V (G) {0, 1,..., p 1} such that the induced function f given by f (uv) = [f(u)] 2 + [f(v)] 2 = {1, p 1, p 2,..., p q 1 } where p i are prime numbers. Then by theorem 4.4.4, there exists a vertex u 1 in V (G) such that f(u 1 ) = 0 and d(u 1 ) = 1. Define a function f 1 : V (G) {1, 2,..., p} such that f 1 (u 1 ) = p and f 1 (u i ) = f(u i ), 1 i p 1. Since d(u 1 ) = 1, the vertex u i which is adjacent to u 1 is such that f(u i ) = 1. Hence f(u i ) and f(u 1 ) are relatively prime. For all the other edges uv E(G), f 1 (u i ) and f 1 (u 1 ) are relatively prime by theorem Hence f 1 is a prime labeling of G so that G is a prime graph. Corollary The converse of the theorem is not always true. Figure 4.11 gives a prime graph which is not prime square sum.

24 Chapter Figure 4.11: Observation for n 2. The complete graph K n is prime square sum Theorem n 2. The star K 1,n is prime square sum if and only if Proof. For n = 1, 2, the prime square sum labeling for K 1,n is given in the figure Let n > 2. By the theorem 4.4.4, if u V (K 1,n ) is such that f(u) = 0,then d(u) = 1. If we label the central vertex of K 1,n as any one of the numbers 1, 2,..., n we get the edge values are not all prime numbers. Hence the star K 1n is not prime square sum for n > 2. We pose the following conjectures and problems: Conjecture Every bipartite graph can be embedded into a prime square sum graph. Conjecture square sum graph. Every tree can be embedded into a prime Problem Characterize the class of prime square sum trees.

25 Chapter References [AG96] S. Arumugam and K.A. Germina. On indexable graphs. Discrete Mathematics, (161): , [AH91a] B.D. Acharya and S. M. Hegde. On certain vertex valuations of graphs. Indian J. Pure Appl Math., 22: , [AH91b] B.D. Acharya and S.M. Hegde. Strongly indexable graphs. Discrete Mathematics, (93): , [BH01] L.W. Beineke and S. M. Hegde. Strongly multiplicative graphs. Discussions Mathematicac, Graph Theory., 21:63 75, [Bur06] [Gal05] [Gol80] David M. Burton, editor. Elementary Number Theory. TATA McGRAW-HILL,, J.A. Gallian. A dynamic survey of graph labelling. The Electronic Journal of Combinatorics(DS6), pages 1 148, Martin Charles Goloumbic. Algorithmic Graph Theory and Perfect Graphs. Academic Press Inc., New York, [Har72] Frank Harary. Graph Theory. Addision Wesley, Massachusetts, [Tel96] S. G. Telang. Number Theory. TATA McGRAW-HILL,, 6 edition, 1996.